AP Chemistry Quiz: Solubility

What this quiz covers

This quiz focuses on Solubility, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Yes; water's polarity allows strong ion–dipole attractions that can overcome the ionic lattice. (correct answer)
2. No; $\mathrm{MgSO_4}$ is insoluble because water's hydrogen bonding prevents ions from separating.
3. No; $\mathrm{MgSO_4}$ is insoluble because sulfates never dissolve in water.
4. Yes; it dissolves because the dissolving rate is fast, which means the solubility must be high.
5. No; ionic solids cannot dissolve in water because water molecules are neutral overall.

Explanation: This question tests knowledge of ionic compound solubility in polar solvents via ion-dipole interactions. Magnesium sulfate is an ionic compound, and water's high polarity enables strong ion-dipole attractions that hydrate and stabilize the Mg2+ and SO4 2- ions. These interactions are sufficient to overcome the lattice energy of MgSO4, allowing the ions to separate and dissolve. Hence, MgSO4 is soluble in water, as the polar solvent effectively surrounds and isolates the ions. Choice A is tempting but wrong, as it misstates that ionic solids cannot dissolve in neutral water molecules, overlooking ion-dipole forces that enable dissolution. For predicting ionic solubility, assess if the solvent can provide stabilizing interactions like ion-dipole to counter the lattice forces.

Question 2

1. Yes; acetone is polar, and polar solutes are most soluble in nonpolar solvents due to dipole alignment.
2. No; acetone is polar while $\mathrm{CCl_4}$ is nonpolar, so solute–solvent attractions are relatively weak. (correct answer)
3. Yes; acetone will hydrogen-bond to $\mathrm{CCl_4}$ through chlorine atoms, increasing solubility.
4. No; acetone is insoluble because it has a higher density than $\mathrm{CCl_4}$ and will sink.
5. Yes; vigorous shaking increases solubility, so acetone will become highly soluble in $\mathrm{CCl_4}$.

Explanation: The skill tested is predicting solubility by matching intermolecular forces between polar and nonpolar substances. Acetone is a polar molecule with a significant dipole moment but no O-H for hydrogen bonding, while CCl4 is nonpolar with only London dispersion forces. The polarity mismatch leads to weak dipole-induced dipole interactions, which are not strong enough to favor mixing over the separate pure substances. Therefore, acetone is not highly soluble in CCl4, as the nonpolar solvent cannot adequately interact with the polar solute. Choice A is a tempting distractor, incorrectly stating that polar solutes are most soluble in nonpolar solvents due to dipole alignment, which confuses the 'like dissolves like' principle. A transferable strategy is to classify both solute and solvent as polar or nonpolar and expect high solubility only when they match.

Question 3

1. Yes; the oxygen in ether can hydrogen-bond strongly to ions and pull the lattice apart.
2. No; ether is not polar enough to provide strong ion–dipole attractions to stabilize $\mathrm{Mg^{2+}}$ and $\mathrm{SO_4^{2-}}$. (correct answer)
3. Yes; any solvent with an oxygen atom will dissolve ionic solids due to covalent bonding.
4. No; ionic compounds dissolve only when the solvent is nonpolar so ions do not recombine.
5. Yes; shaking the mixture increases the dissolving rate, which makes $\mathrm{MgSO_4}$ soluble.

Explanation: The skill being tested is assessing the solubility of ionic compounds in weakly polar solvents lacking strong ion-solvating abilities. Magnesium sulfate (MgSO₄) has high-charge ions requiring strong solvation, but diethyl ether is only slightly polar and cannot donate hydrogen bonds. The weak dipole in ether provides insufficient ion-dipole attractions to stabilize Mg²⁺ and SO₄²⁻ compared to their lattice energy. Therefore, dissolution is unlikely due to inadequate solute-solvent interactions. A tempting distractor is choice A, which overstates ether's ability by claiming its oxygen enables strong hydrogen bonding to ions, ignoring the misconception that any oxygen-containing solvent can solvate ions effectively. Always quantify a solvent's polarity and hydrogen-bonding capacity when predicting ionic solubility to ensure accurate assessments.

Question 4

1. No; gases cannot dissolve in water because water molecules are held too tightly by hydrogen bonding.
2. Yes; $\mathrm{HCl}$ dissolves because bubbling increases the dissolving rate, which increases solubility.
3. Yes; $\mathrm{HCl}$ is polar and can interact strongly with water, and it forms ions that are stabilized by hydration. (correct answer)
4. No; $\mathrm{HCl}$ is nonpolar because it is diatomic, so it cannot dissolve in polar water.
5. No; $\mathrm{HCl}$ cannot dissolve because its molar mass is greater than water's.

Explanation: This question tests understanding of polar gas solubility and ionization in water. Hydrogen chloride is polar and ionizes to H+ and Cl- in water, with ions stabilized by hydration and ion-dipole forces. These strong interactions enhance solubility beyond mere polarity. Thus, HCl is likely soluble in water. Choice A is tempting, incorrectly claiming gases cannot dissolve due to hydrogen bonding tightness, overlooking disruptions by compatible solutes. For gases that ionize, consider both molecular interactions and the stability of resulting ions in the solvent.

Question 5

1. Yes; any molecule with an O–H bond is highly soluble in water due to hydrogen bonding.
2. No; the large nonpolar ring reduces overall polarity, so solubility in water is limited. (correct answer)
3. Yes; aromatic rings are polarizable, so they hydrogen-bond strongly to water.
4. No; benzoic acid is insoluble because it has a higher molar mass than water.
5. Yes; solubility depends mainly on mixing speed, so stirring makes it highly soluble.

Explanation: The skill being tested is analyzing solubility of molecules with both polar and nonpolar regions in polar solvents. Benzoic acid has a polar carboxylic acid group for hydrogen bonding but a large nonpolar aromatic ring that reduces overall polarity. The nonpolar ring leads to weaker interactions with water, limiting solubility despite the polar group. Therefore, benzoic acid is not highly soluble in water at room temperature. Choice A is tempting, overgeneralizing that any O-H bond guarantees high solubility, ignoring the impact of nonpolar portions. A transferable strategy is to consider the overall polarity of the molecule, weighing polar and nonpolar parts when predicting solubility in polar solvents.

Question 6

1. Yes; any ionic compound is highly soluble in water because ion–dipole forces always overcome lattice energy.
2. No; CaCO3 has strong ionic attractions in its lattice, so hydration by water is not sufficient to make it very soluble. (correct answer)
3. Yes; CaCO3 dissolves because carbonate can hydrogen bond strongly with water, breaking the lattice completely.
4. No; CaCO3 is insoluble because it is denser than water, so it cannot dissolve and will always sink.
5. Yes; CaCO3 becomes very soluble if crushed into smaller pieces because particle size increases solubility.

Explanation: This question tests understanding of ionic compound solubility and the balance between lattice energy and hydration energy. Calcium carbonate (CaCO3) has an exceptionally high lattice energy due to the +2 charge on Ca2+ and the -2 charge on CO32-, creating very strong electrostatic attractions in the solid. While water can form ion-dipole interactions with the ions, the hydration energy gained is insufficient to overcome the large lattice energy, resulting in very low solubility (Ksp ≈ 10^-9). The carbonate ion's large size and charge distribution also make it less effectively hydrated than smaller, simpler ions. Choice A incorrectly assumes all ionic compounds are highly soluble, ignoring that compounds with high lattice energies (like those with multiply charged ions) often have low solubility. The key insight is that solubility depends on the balance between lattice energy and hydration energy, not simply on whether a compound is ionic.

Question 7

1. Yes; CO2 is highly soluble because nonpolar gases always dissolve well in polar solvents.
2. No; CO2 is nonpolar overall, so it has limited favorable interactions with water compared with water–water hydrogen bonding. (correct answer)
3. Yes; CO2 dissolves because water forms strong hydrogen bonds directly to the carbon atom in CO2.
4. No; CO2 cannot dissolve because gases are never soluble in liquids under any conditions.
5. Yes; CO2 becomes highly soluble if stirred because stirring increases the solubility of gases in liquids.

Explanation: This question tests understanding of gas solubility in polar solvents. Carbon dioxide (CO2) is a linear molecule that, despite having polar C=O bonds, is nonpolar overall due to its symmetry where the bond dipoles cancel out. CO2 interacts primarily through weak London dispersion forces, which provide limited stabilization when dissolved in highly polar, hydrogen-bonding water. The weak CO2-water interactions cannot effectively compete with the strong water-water hydrogen bonds that must be disrupted to accommodate the gas molecules. Choice C incorrectly suggests water forms hydrogen bonds to carbon, but carbon lacks the high electronegativity and lone pairs needed for hydrogen bonding. The strategy is to consider molecular geometry when assessing polarity - linear molecules with identical terminal atoms like CO2 are nonpolar regardless of bond polarity, leading to low solubility in polar solvents.

Question 8

1. Yes; glucose can form many hydrogen bonds with water, leading to strong solute–solvent interactions. (correct answer)
2. No; glucose is a covalent compound, and covalent solutes are always insoluble in water.
3. No; glucose will not dissolve because it is a solid at room temperature, and solids cannot be soluble in liquids.
4. Yes; glucose dissolves because its large molar mass increases its solubility in water.
5. No; glucose will dissolve only if stirred fast enough since stirring changes the equilibrium solubility.

Explanation: This question tests understanding of how molecular structure affects solubility in polar solvents. Glucose (C6H12O6) contains five hydroxyl (-OH) groups that can form extensive hydrogen bonds with water molecules, making it highly polar despite being a molecular compound. Each glucose molecule can form multiple hydrogen bonds as both donor and acceptor with surrounding water molecules, creating a stable hydration shell that makes dissolution energetically favorable. The numerous OH groups provide sufficient solute-solvent interactions to overcome the glucose-glucose interactions in the solid crystal. Choice B incorrectly assumes covalent compounds cannot dissolve in water, failing to recognize that polarity and hydrogen bonding capability, not bond type, determine solubility. The strategy is to count hydrogen bonding sites - molecules with multiple OH groups typically show high water solubility.

Question 9

1. Yes; any ionic compound dissolves completely in water because water is polar.
2. No; strong ionic attractions in the solid (high lattice energy) are not fully overcome by hydration, so solubility is low. (correct answer)
3. Yes; carbonate ions hydrogen-bond strongly with water, making $\mathrm{CaCO_3}$ highly soluble.
4. Yes; stirring increases the solubility of $\mathrm{CaCO_3}$ until all of it dissolves.
5. No; $\mathrm{CaCO_3}$ is insoluble because calcium is a metal and metals cannot dissolve in water.

Explanation: This question tests understanding of how lattice energy affects ionic solubility. Calcium carbonate has exceptionally high lattice energy due to the small, highly charged Ca²⁺ and CO₃²⁻ ions, which create very strong electrostatic attractions in the solid. While water can provide ion-dipole interactions for hydration, the hydration energy is insufficient to overcome CaCO₃'s large lattice energy, resulting in very low solubility (Ksp ≈ 10⁻⁹). This explains why limestone and chalk (forms of CaCO₃) persist in nature despite water exposure. Choice A incorrectly assumes all ionic compounds are highly soluble in water, but solubility depends on the balance between lattice energy and hydration energy—compounds with high lattice energies relative to hydration energies have low solubility. When predicting ionic solubility, consider both ion charges and sizes: small, highly charged ions typically form less soluble compounds.

Question 10

1. No; acetone cannot dissolve in water because it lacks an O–H bond and therefore cannot interact with water.
2. Yes; acetone is polar and can engage in dipole interactions and hydrogen bonding (as an acceptor) with water. (correct answer)
3. No; acetone is soluble only in nonpolar solvents because its methyl groups make it entirely nonpolar.
4. Yes; acetone will be miscible only if the mixture is stirred, since stirring determines miscibility.
5. No; acetone will not dissolve because its boiling point is lower than water's, preventing mixing.

Explanation: This question tests understanding of polar molecular solubility and hydrogen bonding interactions. Acetone ((CH3)2CO) has a polar carbonyl group (C=O) with a significant dipole moment, making the molecule polar overall despite having two methyl groups. The oxygen atom in acetone has lone pairs that can accept hydrogen bonds from water molecules, while water can also interact with acetone through dipole-dipole forces. These favorable interactions allow acetone and water to be completely miscible, as the acetone-water interactions effectively replace water-water hydrogen bonds. Choice A incorrectly claims acetone cannot interact with water due to lacking an O-H bond, confusing hydrogen bond donation with hydrogen bond acceptance. The strategy is to identify polar functional groups and remember that molecules can participate in hydrogen bonding as acceptors even without O-H, N-H, or F-H bonds.

Question 11

1. Yes; MgCl2 is ionic and water is polar, so ion–dipole interactions can stabilize Mg2+ and Cl− in solution. (correct answer)
2. No; MgCl2 is ionic, and ionic compounds dissolve only in nonpolar solvents through dispersion forces.
3. No; MgCl2 will not dissolve because Mg2+ has too high a charge, and higher charge always decreases solubility to zero.
4. Yes; MgCl2 dissolves only if heated because temperature is the primary factor determining whether dissolution is possible.
5. No; MgCl2 will not dissolve because it is a solid, and only liquids can be soluble in water.

Explanation: This question tests understanding of ionic compound solubility in polar solvents. Magnesium chloride (MgCl2) is an ionic compound that readily dissolves in water because the polar water molecules can form strong ion-dipole interactions with both Mg2+ and Cl- ions. The hydration energy from water molecules surrounding the ions is sufficient to overcome the lattice energy of MgCl2, making dissolution thermodynamically favorable. Water's high dielectric constant also helps stabilize the separated ions by reducing the electrostatic attraction between them. Choice B incorrectly states that ionic compounds dissolve only in nonpolar solvents, which contradicts the fundamental principle that ionic compounds require polar solvents for dissolution. The key concept is that ionic compounds generally dissolve well in polar solvents through ion-dipole interactions, with water being particularly effective due to its high polarity.

Question 12

1. No; benzene is nonpolar, so it can dissolve only ionic compounds and not molecular solids.
2. Yes; both are nonpolar and interact primarily via London dispersion forces, so mixing is favorable. (correct answer)
3. No; naphthalene has a higher melting point, so it cannot dissolve in any liquid solvent.
4. Yes; heating increases the rate of dissolving, which proves that naphthalene is soluble in benzene.
5. No; benzene molecules are symmetrical, so they cannot attract solute particles strongly enough to dissolve them.

Explanation: This question tests understanding of solubility based on matching nonpolar characteristics. Both naphthalene and benzene are nonpolar aromatic compounds with delocalized π-electron systems, interacting primarily through London dispersion forces. According to "like dissolves like," nonpolar solutes readily dissolve in nonpolar solvents because the intermolecular forces are compatible—naphthalene molecules can easily integrate into the benzene liquid without disrupting existing interactions. The similar molecular structures (both containing aromatic rings) further enhance compatibility. Choice A incorrectly states that nonpolar solvents dissolve only ionic compounds, which is backwards—nonpolar solvents actually cannot dissolve ionic compounds due to inability to stabilize separated ions. To predict solubility between organic compounds, identify if both are polar or nonpolar based on molecular structure and symmetry.

Question 13

1. Yes; both are largely nonpolar and interact mainly through dispersion forces, so mixing is favorable. (correct answer)
2. No; nonpolar solutes require polar solvents to induce dipoles and pull them into solution.
3. Yes; toluene dissolves naphthalene because it can hydrogen-bond to the aromatic ring.
4. No; naphthalene is insoluble because its melting point is high, which prevents dissolving.
5. No; naphthalene dissolves too slowly in toluene at room temperature, so it is not soluble.

Explanation: The skill being assessed is evaluating solubility of nonpolar substances in nonpolar solvents using dispersion forces. Naphthalene is a nonpolar aromatic hydrocarbon relying on London dispersion forces, similar to toluene, which is also nonpolar and aromatic. This compatibility allows strong dispersion force interactions between solute and solvent, facilitating dissolution. Therefore, naphthalene is likely soluble in toluene, as both share the same type of intermolecular forces. Choice B is a tempting distractor, incorrectly claiming nonpolar solutes need polar solvents to induce dipoles, which contradicts the 'like dissolves like' rule. To solve solubility questions, identify shared intermolecular force types between solute and solvent for favorable outcomes.

Question 14

1. Yes; sucrose can form hydrogen bonds with hexane, so the solute–solvent attractions are strong.
2. No; sucrose is highly polar and hydrogen-bonding, while hexane is nonpolar, so solute–solvent IMFs are too weak. (correct answer)
3. Yes; hexane has a low boiling point, so it dissolves solids more effectively than high–boiling point solvents.
4. No; sucrose dissolves slowly in hexane because hexane molecules are too large to surround the solute quickly.
5. Yes; any solute will dissolve if enough stirring is provided, because stirring increases solubility.

Explanation: This question tests the skill of predicting solubility based on the compatibility of intermolecular forces between solute and solvent using the 'like dissolves like' principle. Sucrose is a highly polar molecule with multiple O-H groups that enable strong hydrogen bonding, while hexane is nonpolar and relies solely on London dispersion forces. The mismatch in polarity means that the solute-solvent interactions are weak dispersion forces, which are insufficient to overcome the strong hydrogen bonds in sucrose and the dispersion forces in hexane. Therefore, sucrose is not likely to dissolve well in hexane, as the nonpolar solvent cannot effectively stabilize the polar solute molecules. A tempting distractor is choice A, which incorrectly assumes sucrose can form hydrogen bonds with hexane, misunderstanding that hydrogen bonding requires compatible polar groups like O-H or N-H in both solute and solvent. To predict solubility effectively, always compare the dominant intermolecular forces of the solute and solvent to ensure they can form strong attractive interactions.

Question 15

1. Yes; any polar solvent will fully dissolve any ionic solid due to dipole–dipole forces.
2. No; acetone's polarity may not provide sufficient ion–dipole stabilization to overcome the ionic lattice. (correct answer)
3. Yes; acetone donates hydrogen bonds to separate $\mathrm{K^+}$ and $\mathrm{I^-}$ ions.
4. No; $\mathrm{KI}$ is insoluble because iodide ions are nonpolar and cannot dissolve in any solvent.
5. Yes; increasing temperature always guarantees complete solubility of ionic solids in any solvent.

Explanation: This question evaluates ionic solubility in polar aprotic solvents. Potassium iodide is ionic, and acetone is moderately polar but cannot donate hydrogen bonds, providing weaker ion-dipole stabilization. This may not suffice to overcome KI's lattice energy, leading to limited solubility. Thus, KI is not expected to be very soluble in acetone. Choice A is a tempting distractor, overgeneralizing that any polar solvent fully dissolves any ionic solid, neglecting solvent-specific properties like hydrogen-bonding ability. For ionic compounds, assess the solvent's polarity and hydrogen-bonding capacity to determine if it can effectively hydrate ions.

Question 16

1. No; ethylene glycol has covalent bonds, and covalent substances do not dissolve in water.
2. Yes; both substances are polar and can hydrogen-bond, so strong solute–solvent attractions favor mixing. (correct answer)
3. No; water has stronger hydrogen bonding than ethylene glycol, so water will not allow it into solution.
4. Yes; ethylene glycol dissolves because its molar mass is smaller than water's.
5. No; ethylene glycol dissolves too slowly at room temperature, so it is considered insoluble.

Explanation: The skill being tested is determining solubility through analysis of molecular polarity and hydrogen-bonding capabilities. Ethylene glycol is a polar molecule with two O-H groups that can both donate and accept hydrogen bonds, matching water's polar and hydrogen-bonding nature. This similarity allows for strong solute-solvent hydrogen bonds and dipole-dipole interactions, which overcome the intermolecular forces within each pure substance. As a result, ethylene glycol is highly soluble in water, forming a homogeneous solution due to these favorable attractions. Choice A is incorrect and tempting as it confuses covalent bonding with solubility rules, mistakenly believing covalent compounds do not dissolve in water despite many polar covalent molecules doing so. When assessing solubility, evaluate if the solute can participate in the same types of intermolecular forces as the solvent to promote effective dissolution.

Question 17

1. No; a nonpolar solute has weak attractions to polar water, so dissolution is unfavorable. (correct answer)
2. No; naphthalene is insoluble because it has a high melting point, not because of polarity.
3. Yes; water dissolves all molecular solids because it can hydrogen-bond with any solute.
4. Yes; crushing the solid would increase surface area, which makes it soluble in water.
5. Yes; nonpolar solutes dissolve best in polar solvents because dipoles induce strong bonding.

Explanation: The skill being tested is predicting insolubility of nonpolar solutes in polar solvents due to mismatched intermolecular forces. Naphthalene is nonpolar, interacting via London dispersion forces, while water is polar with strong hydrogen bonding. The 'like dissolves like' principle indicates that the weak attractions between naphthalene and water cannot overcome water's hydrogen bond network or naphthalene's dispersion forces. Thus, little dissolving occurs, as observed, because the process is energetically unfavorable. A tempting distractor is choice A, which wrongly suggests water dissolves all molecular solids via hydrogen bonding, based on the misconception that water's versatility extends to nonpolar substances. To avoid errors, systematically compare the dominant intermolecular forces of solute and solvent before predicting solubility outcomes.

Question 18

1. Yes; water's polarity allows ion–dipole attractions that can stabilize separated $\mathrm{Ca^{2+}}$ and $\mathrm{Cl^-}$ ions. (correct answer)
2. No; ionic compounds only dissolve in nonpolar solvents because charges repel polar molecules.
3. No; $\mathrm{CaCl_2}$ has strong ionic bonds, so it cannot dissolve in any solvent.
4. Yes; increasing stirring speed increases the reaction rate, which makes $\mathrm{CaCl_2}$ soluble.
5. No; water forms hydrogen bonds, so it can only dissolve substances that hydrogen-bond directly.

Explanation: The skill being tested is determining the solubility of ionic compounds in polar solvents through ion-dipole interactions. Calcium chloride consists of Ca²⁺ and Cl⁻ ions held by strong ionic bonds, but water's polarity allows for ion-dipole attractions that can stabilize these ions in solution. The process of dissolution involves water molecules surrounding and hydrating the ions, overcoming the lattice energy of the solid. Thus, CaCl₂ is likely soluble in water due to these favorable interactions. A tempting distractor is choice C, which wrongly claims ionic bonds prevent dissolution in any solvent, reflecting the misconception that lattice strength alone determines solubility without considering solvent interactions. A transferable strategy is to consider both the strength of solute-solute attractions and the potential for solute-solvent attractions when assessing solubility.

Question 19

1. No; $\mathrm{I_2}$ is nonpolar and therefore can only dissolve in polar solvents like water.
2. No; $\mathrm{CCl_4}$ has polar bonds, so it behaves as a polar solvent and repels $\mathrm{I_2}$.
3. Yes; both are nonpolar, so dispersion forces between solute and solvent can be favorable. (correct answer)
4. No; $\mathrm{I_2}$ dissolves too slowly at room temperature, so it is insoluble in $\mathrm{CCl_4}$.
5. Yes; dissolving occurs because $\mathrm{I_2}$ chemically reacts with $\mathrm{CCl_4}$ to form ions.

Explanation: The skill being tested is evaluating solubility of nonpolar substances in nonpolar solvents using London dispersion forces. Iodine (I₂) is a nonpolar molecule where particles interact via dispersion forces, and carbon tetrachloride (CCl₄) is similarly nonpolar. The 'like dissolves like' rule applies here, as both can form comparable dispersion interactions, making dissolution favorable. No chemical reaction is needed; the similarity in forces allows I₂ to mix into CCl₄. A tempting distractor is choice A, which mistakenly states nonpolar substances dissolve only in polar solvents, based on the misconception that opposites attract in solubility rather than similarities. Remember, to assess nonpolar solubility, check if both solute and solvent rely primarily on dispersion forces for compatibility.

Question 20

1. Yes; gases dissolve best in water because water has empty space between molecules.
2. No; propane is nonpolar and cannot form strong attractions with polar water, so solubility is low. (correct answer)
3. Yes; propane can hydrogen-bond with water through its C–H bonds.
4. No; propane is insoluble because it reacts with water to form alcohols.
5. Yes; shaking the container increases solubility, so propane becomes highly soluble in water.

Explanation: The skill tested is predicting gas solubility in liquids based on polarity and intermolecular forces. Propane is a nonpolar gas with London dispersion forces, while water is polar with hydrogen bonding, leading to weak solute-solvent interactions. These weak forces cannot compete with water's strong hydrogen bonds, resulting in low solubility for propane. Therefore, propane is not very soluble in water at room temperature. Choice C is a tempting distractor, mistakenly suggesting C-H bonds in propane can hydrogen-bond with water, confusing weak C-H with true hydrogen bonding requiring O-H, N-H, or F-H. When evaluating gas solubility, consider if the gas molecules can form strong enough attractions with the solvent to disrupt its structure.